Definite Integral As Limit Of A Sum And Estimate Of Definite Integrals

# Definite Integral As Limit Of A Sum And Estimate Of Definite Integrals | Mathematics (Maths) for JEE Main & Advanced PDF Download

## Definite Integral As Limit Of A Sum

Remark :

The symbol   was introduced by Leibnitz and is called integral sign. It is an elongated S and was chosen because an integral is a limit of sums. In the notation    is called the integrand and a and b are called the limits of integration; a is the lower limit and b is the upper limit. The symbol dx has no official meaning by itself;    is all one symbol. The procedure of calculating an integral is called integration.

## Estimate Of Definite Integration & General Inequality

STATEMENT : If f is continuous on the interval [a, b], there is atleast one number c between a and b such that

Proof : Suppose M and m are the largest and smallest values of f, respectively, on [a, b]. This means m ≤ f(x) ≤ M when  a ≤ x ≤ b

Because f is continuous on the closed interval [a, b] and because the number I =  lies between m and M, the intermediate value theorem syas there exists a number c between a and b for which f(c) = I ; that is,

The mean value theorem for integrals does not specify how to determine c. It simply guarantees the existence of atleast one number c in the interval.
Since f(x) = 1 + x2 is continuous on the interval [–1, 2], the Mean Value Theorem for Integrals says there is a number c in [–1, 2] such that

In this particular case we can find c explicitly. From previous Example we know that fave = 2, so the value of c satisfies f(c) =  = 2

Thus, in this case there happen to be two numbers c = ± 1 in the interval [–1, 2] that work in the mean value theorem for Integrals.

Walli's Formula & Reduction Formula

## Solved Examples

Ex.1 Evaluate   using limit of sum.

Sol.

This integral can't be interpreated as an area because f takes on both positive and negative values. But is can be interpreated as the difference of areas A1 – A2, where A1 and Aare shown in Figure

Ex.2 Prove that,   Hence or otherwise evaluate

Sol.

Consider sin nθ + sin (n - 2) θ = 2 sin (n - 1) θ cos θ ⇒ sin nθ sec θ = 2 sin (n-1) θ - sin (n - 2) θ sec θ

Hence

,

Ex.3 Prove that

Sol.

L.H.S. =

(From Walli's formula)

Ex.4 If un =     then show that u1, u2, u3 ,...... constitute an arithmetic progression. Hence or otherwise find the value of un.

Sol.

un + 1 – 2un + un – 1 = (un + 1 – un) – (un – un – 1)

∴ un – 1 + un + 1 = 2ui.e., un – 1, un, un + 1 form an A.P.

⇒ u1, u2, u3,.........constitute an A.P.
Ex.5 Evaluate

Sol.

Integrating by parts taking unity as the second function, we have

Ex.6 Show that    Hence or otherwise evaluate

Sol.

Put a tan x = t ⇒ a sec2x dx = dt when x = 0 ⇒ t = 0 ;  x = p/2 ⇒ t = ∝

Differentiating both side w.r.t. 'a', we get

again differentiating both sides w.r.t. 'a' we get

Put a = √5 on both sides, we get

Ex.7 Let f be an injective functions such that f(x) f(y) + 2 = f(x) + f(y) + f(xy) for all non negative real x and y with f(0) = 1 and f '(1) = 2 find f(x) and show that f(x) dx – x (f(x) + 2) is a constant.

Sol. We have f(x) f(y) + 2 = f(x) + f(y) + f(xy)                          ......(1)
Putting x = 1 and y = 1 then f(1) f(1) + 2 = 3f(1)
we get f(1) = 1, 2 & f(1)  1           (  f(0) = 1 & function is injective) then f(1) = 2

Replacing y by 1/x in (1) then f(x) f(1/x) + 2 = f(x) + f(1/x) + f(1) ⇒  f(x) f(1/x) = f(x) + f(1/x) [ f(1) = 2)

Hence f(x) is of the type f(x) = 1 ± xn ⇒ f(1) = 1 ± 1 = 2 (given)

∴ f(x) = 1 + xn  and f '(x) = nxn – 1 ⇒ f '(1) = n = 2 ∴ f(x) = 1 + x2

∴

Ex.8 Evaluate

Sol.

Let  I =  Make |sin x| – |cos x| = 0 ∴  |tan x| = 1

∴ tan x = ± 1    and both these values lie in the interval [0, π].

We find for 0 < x < π/4, |sin x| – |cos x| < 0

|sin x| – |cos x| > 0

Ex.9 Evaluate  , (where [ * ] is the greatest integer function)

Sol. Let I =

Let f(x) = x2 + x + 1  ⇒  f '(x) = 2x - 1 for x > 1/2, f '(x) > 0 and x < 12, f '(x) , 0

Values of f(x) at x = 1/2 and 2 are 3/4 and 3 integers between them an 1, 2 then x2 - x + 1 = 1, 2

we get x = 1,   and values of f(x) at x = 0 and 1/2 are 1 and 3/4 no integer between them

Alternative Method :  It is clear from the figure

Ex.10 If

Sol.

Integrating by parts taking x2 as 1 st function, we we get  =

[By Prop.]

Adding (1) and (2) we get

Ex.11 Prove that

Sol.

Intergrating by parts taking x as a first function, we have

Ex.12 Evaluate

Sol.

Let g(t) = |t – 1| – |t| + |t + 1| =

=

Ex.13 For all positive integer k, prove that

Hence prove that

Sol. We have 2 sinx [cos x + cos 3x + ...+ cos (2k–1)x ]
= 2 sinx cosx + 2 sinx cos 3x + ....+ 2 sinx cos(2k – 1)x
= sin 2x + sin 4x – sin 2x + sin 6x – sin 4x + ....... + sin 2kx – sin (2k – 2) x
= sin 2kx
⇒ 2[cos x + cos 3x + ........ + cos (2k – 1) x ] = sin2kx/sinx

[2 cos2x + 2 cos 3x cos x + ....... + 2 cos (2k – 1 ) x cosx ] dx

[cosx + cos 3x + ....... + cos (2k – 1 ) x ]cosx dx

=

Ex.14 Let f(x) is periodic function such that  Find the function f(x) if (1) = 1 .

Sol.

from (2) and (3)

Differentiating bot sides w.r.t.x, we get

then f(x) = 1/x3 But given f(x) is a periodic function Hence f(x) = 1

Ex.15 Assume   then prove that

Sol.

Ex.16 Use induction to prove that ,

Sol.

Now sinkx = sin[(k + 1) x – x ] = sin(k + 1) x cos – cos(k + 1)x sin x

Hence sin (k + 1 ) x cosx = sinkx + cos(k + 1) x sin x

Subistuting P(k + 1) =

Now I. B. P. to get the result

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## FAQs on Definite Integral As Limit Of A Sum And Estimate Of Definite Integrals - Mathematics (Maths) for JEE Main & Advanced

 1. What is the concept of definite integral as a limit of a sum?
The concept of definite integral as a limit of a sum is a fundamental idea in calculus. It involves dividing a region under a curve into smaller and smaller intervals, and then taking the limit as the interval size approaches zero. This process allows us to find the exact area under a curve between two given points.
 2. How can we estimate definite integrals?
Definite integrals can be estimated using various numerical methods, such as the midpoint rule, trapezoidal rule, or Simpson's rule. These methods involve dividing the interval over which the integral is being computed into smaller subintervals and approximating the area under the curve using geometric shapes.
 3. Can definite integrals be used to find the value of a function at a specific point?
No, definite integrals cannot be used to find the value of a function at a specific point. Definite integrals give us the net area under a curve between two points, but they do not provide information about the function's value at any particular point. To find the value of a function at a specific point, we need to use the concept of the antiderivative or evaluate the function directly.
 4. What is a general inequality related to definite integrals?
One general inequality related to definite integrals is the Mean Value Theorem for Integrals. It states that if a function f(x) is continuous on the closed interval [a, b] and integrable on the open interval (a, b), then there exists a number c in (a, b) such that the definite integral of f(x) over the interval [a, b] is equal to f(c) times the length of the interval (b-a). Symbolically, ∫[a,b] f(x) dx = f(c) * (b-a).
 5. How are definite integrals used in real-world applications?
Definite integrals are used in various real-world applications, such as calculating areas and volumes, finding average values, determining total accumulated quantities, and solving problems involving rates of change. They are widely used in physics, engineering, economics, and many other fields where the concept of accumulation or total is relevant.

## Mathematics (Maths) for JEE Main & Advanced

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